Help with a Real Analysis Proof

[vyro]

Homework Statement

Prove that 2^n + 3^n is a multiple of 5 for all odd n that exist in the set of natural numbers.

Homework Equations

The Attempt at a Solution

Suppose the contrary perhaps and do a proof by contradiction? Perhaps induction?

edit: done, thank you. please look at second proof :)

 

[vyro]

Also, another proof:

Homework Statement

Prove that for each a that exists in real numbers and n that exists in natural numbers there exists a rational r such that abs(a-r) < 1/n

Homework Equations

The Attempt at a Solution

I believe it has something to do with the theorem on density of rationals.

 

[boboYO]

i don't think the first question is a real analysis problem. try working in mod 5. what's 2^n mod 5 and 3^n mod 5 for odd n?

 

[vyro]

yeah, i redid the first one using a proof by induction. thanks for the response though.

 

[HallsofIvy]

Also, another proof:

Homework Statement

Prove that for each a that exists in real numbers and n that exists in natural numbers there exists a rational r such that abs(a-r) < 1/n

Homework Equations

The Attempt at a Solution

I believe it has something to do with the theorem on density of rationals.

Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).

 

[vyro]

Yes, saying that |a- r|= |r- a|< 1/n is the same as saying that -1/n< r- a< 1/n or that a- 1/n< r< a+ 1/n. Since there exist a rational number in any interval, there exist a rational number in the interval (a- 1/n, a+ 1/n).

is that valid as a proof though, starting from the conclusion and then reaffirming it?